Find the general solution for 3sinx+4cosx=5

Steps to solve Find the general solution for 3sinx+4cosx=5

Put 3=rcosθ and 4=rsinθ

⇒3^2+4^2=r^2.cos^2θ+r^2.sin^2θ

⇒25=r^2(cos^2θ+sin^2θ)

⇒r^2=25

⇒r=±5

Again

⇒(4/3)=(rsinθ/rcosθ)

⇒tanθ=(4/3)

⇒θ=tan^(-1)(4/3)

From question

⇒rcosθ.sinx+rsinθ.cosx=5

⇒r(cosθ.sinx+sinθ.cosx)=5

⇒5sin(x+θ)=5

⇒sin(x+θ)=1

⇒sin(x+θ)=sin(π/2)

⇒x+θ=π/2

⇒x=nπ+(-1)^n(π/2)-tan^(-1)(4/3)

Hence the general solution for 3sinx+4cosx=5 is nπ+(-1)^n(π/2)-tan^(-1)(4/3)

Detail Information:-

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